Answer:
If the pressure is increased, the equilibrium position moves in the direction of the fewest molecules of gas. This means it moves to the right in the Haber process. ... Stronger equipment is needed, and more energy is needed to compress the gases. So a compromise pressure of 200 atmospheres is chosen.
Explanation:
I think it’s b I’m not sure so sorry if incorrect. Take care.
<span>C) the core is more than a million times hotter than the surface</span>
It stays roughly the same size. Electrons have a barely imperceptible mass so the overall mass of the atom is changed very little.
Answer:
A and C are true , B and D are false
Explanation:
For A)
from the first law of thermodynamics (in differential form)
dU= δQ - δW = δQ - PdV
from the second law
dS ≥ δQ/T
then
dU ≤ T*dS - p*dV
dU - T*dS + p*dV ≤ 0
from the definition of Gibbs free energy
G=H - TS = U+ PV - TS → dG= dU + p*dV + V*dp - T*dS - S*dT
dG - V*dp + S*dT = dU - T*dS + p*dV ≤ 0
dG ≤ V*dp - S*dT
in equilibrium, pressure and temperature remains constant ( dp=0 and dT=0). Thus
dG ≤ 0
ΔG ≤ 0
therefore the gibbs free energy should decrease in an spontaneous process → A reaction with a negative Gibbs standard free energy is thermodynamically spontaneous under standard conditions
For B) Since the standard reduction potential is related with the Gibbs standard free energy through:
ΔG⁰=-n*F*E⁰
then, when ΔG⁰ is negative , E⁰ is positive and therefore a coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous.